ZA-SAMO 2011 Problem 6

In triangle $ABC$ , the incircle touches $BC$ in $D$ , $CA$ in $E$ and $AB$ in $F$ . The bisector of $\angle BAC$ intersects $BC$ in $G$ . The lines $BE$ and $CF$ intersect in $J$ . The line through $J$ perpendicular to $EF$ intersects $BC$ in $K$ . Prove that $\frac{GK}{DK}=\frac{AE}{CE}+\frac{AF}{BF}$